| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 109702 | Accepted: 34255 |
Description
Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting.
* Walking: FJ can move from any point X to the points X - 1 or X + 1 in a single minute
* Teleporting: FJ can move from any point X to the point 2 × X in a single minute.
If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?
Input
Line 1: Two space-separated integers: N and K
Output
Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.
Sample Input
5 17
Sample Output
4
Hint
The fastest way for Farmer John to reach the fugitive cow is to move along the following path: 5-10-9-18-17, which takes 4 minutes.
Source
思路：简单的宽搜，但是改了很长时间。
求大佬指正：这里哪儿有问题呢？
void bfs(){
while(!que.empty()) que.pop();
nond tmp;tmp.pos=x;tmp.step=;
vis[x]=;que.push(tmp);
while(!que.empty()){
nond now=que.front();
que.pop();nond a,b,c;
c.pos=now.pos+;c.step=now.step+;if(now.pos<=y&&!vis[c.pos]) que.push(c),vis[c.pos]=;
if(c.pos==y){ printf("%d\n",c.step);return ; }
a.pos=now.pos-;a.step=now.step+;if(now.pos>=&&!vis[a.pos]) que.push(a),vis[a.pos]=;
if(a.pos==y){ printf("%d\n",a.step);return ; }
b.pos=now.pos*;b.step=now.step+;if(now.pos<=y&&!vis[b.pos]) que.push(b),vis[b.pos]=;
if(b.pos==y){ printf("%d\n",b.step);return ; }
}
}
像上面那样写就不行，改成下面这样就AC了。
#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int x,y;
int vis[];
struct nond{ int pos,step; };
queue<nond>que;
void bfs(){
while(!que.empty()) que.pop();
nond tmp;tmp.pos=x;tmp.step=;
vis[x]=;que.push(tmp);
while(!que.empty()){
nond now=que.front();
que.pop();nond a,b,c;
if(now.pos==y){ printf("%d\n",now.step);return ; }
c.pos=now.pos+;c.step=now.step+;if(now.pos<=y&&!vis[c.pos]) que.push(c),vis[c.pos]=;
a.pos=now.pos-;a.step=now.step+;if(now.pos>=&&!vis[a.pos]) que.push(a),vis[a.pos]=;
b.pos=now.pos*;b.step=now.step+;if(now.pos<=y&&!vis[b.pos]) que.push(b),vis[b.pos]=;
}
}
int main(){
while(scanf("%d%d",&x,&y)!=EOF){
memset(vis,,sizeof(vis));
bfs();
}
}